WAEC - Further Mathematics (2011 - No. 14)
If \(y^{2} + xy - x = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\).
\(\frac{1 - y}{2y}\)
\(\frac{1 - 2y}{x}\)
\(\frac{1 - y}{x + 2y}\)
\(\frac{1}{x + 2y}\)
Explanation
Given \(y^{2} + xy - x = 0\)
Using the method of implicit differentiation, we have
\(2y\frac{\mathrm d y}{\mathrm d x} + x\frac{\mathrm d y}{\mathrm d x} + y - 1 = 0\)
\(\frac{\mathrm d y}{\mathrm d x}(2y + x) = 1 - y\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{1 - y}{x + 2y}\)
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